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Problem: Is there a polynomial $p(x)$ with integer coefficients such that $p(2013)=1789$ and $p(1515)=1830$?

My attempt:

After ruling out polynomials of degree 1, 2 and 3, and with further inspection, it appeared to me that such a polynomial doesn't exist.

Is my below proof accurate?

Suppose that $p$ exists, and let

$q(x)=p(x)-1789$ and $s(x)=p(x)-1830$

then, $q(2013)=0$ and $s(1515)=0$, so clearly,

$q(x)=(x-2013)f(x)$ and $s(x)=(x-1515)g(x)$,

for some polynomials $f$ and $g$ with integer coefficients.

Let $a$, $c$ and $d$ be the constant terms of $p$, $g$ and $f$, respectively, and observe that the constant terms of $s$ and $q$ are

$-1515c=a-1830 \ $ and $-2013d=a-1789$, $ \ $ respectively. $ \ \ \ \ \ \ (*)$

Thus,

$a \equiv 1830 \equiv 315 \pmod{1515}$ and $a \equiv 1789 \pmod{2013}$.

But this system of congruence doesn't have a simultaneous solution (since $ \gcd(1515, 2013)\not\mid 1789-315$).

Hence, there's no integer value of $a$ that satisfies $(*)$, which is a contradiction.
Therefore, $p$ doesn't exist.

If the above is correct, is there perhaps a more direct way of proving this?

Michael Rozenberg
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Stephen
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    $a-b$ divides $P(a)-P(b)$, when $a,b$ integers and $P$ integral polynomial and that is manifestly not the case here – Conrad Jul 10 '19 at 18:41
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    From $p(1515) = 1830$ you know that the constant coefficient $a_0$ must be divisible by $3$. However, in that case $p(2013)$ would also be divisible by $3$ so it cannot equal $1789$. – Tob Ernack Jul 10 '19 at 18:42
  • because $1515$ and $1830$ are divisible by $3$ – J. W. Tanner Jul 10 '19 at 18:49
  • @TobErnack Got it now. Clearly shorter! – Stephen Jul 10 '19 at 18:49
  • Similarly with mod 2, $2013 \equiv 1515 \pmod{2}$ but $1789 \not\equiv 1830 \pmod{2}$. – Daniel Schepler Jul 10 '19 at 19:24
  • @Bill Dubuque This is not a duplicate. It's not even the same question as the one you linked...What are you on about? – Stephen Jul 10 '19 at 20:40
  • @Ste It's most certainly an abstract duplicate Problems like this follow easily (here immediately) from the Factor theorem $,a -b\mid P(a)- P(b),$ as explained in the linked dupes. We probably have tens if not hundreds of these dupes. It is unlikely that anything novel will be added in this very simple instance (which I'm 100% sure already occurs multiple times here but with different numbers). Please consider retracting your reopen vote to help curtail this rampant duplication. – Bill Dubuque Jul 10 '19 at 20:47
  • @BillDubuque Did you even read my question? 1) not only is it different from the ones you linked, but 2) I'm also asking whether my attempt is correct, and how it could be made simpler. – Stephen Jul 10 '19 at 20:47
  • @Stephen But you accepted an answer that says absolutely nothing about that (and the answer is a dupe). – Bill Dubuque Jul 10 '19 at 20:48
  • @BillDubuque I deduced an appropriate answer to my question from the comments, and that answer below - so I decided to accept it. What's wrong with that? – Stephen Jul 10 '19 at 20:50
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    I don't agree that it's duplicate. By this way, all math problems they are duplicates. I opened this topic. – Michael Rozenberg Jul 11 '19 at 00:09

1 Answers1

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$p(2013)-p(1515)$ is divisible by $2013-1515=498,$ but $1789-1830=-41$ is not.

It follows from the following reasoning.

Let $p(x)=a_0x^n+a_1x^{n-1}+...+a_n,$ where $a_i\in\mathbb Z$.

Thus, $$p(m)-p(k)=a_0(m^n-k^n)+a_1(m^{n-1}-k^{n-1})+...+a_{n-1}(m-k)=$$ $$=(m-k)(a_0(m^{n-1}+m^{n-2}k+...+k^{n-1})+a_1(m^{n-2}+...+k^{n-2})+...+a_{n-1}).$$

Michael Rozenberg
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